Question

find the value of the sum \\(\sum\limits_{j=1}^{32} (4j^2 - 8j + 2)\\). use a summation formula, if possible.

Explanation:

Step1: Split the summation

$$\sum_{j=1}^{32} (4j^2 - 8j + 2) = 4\sum_{j=1}^{32} j^2 - 8\sum_{j=1}^{32} j + \sum_{j=1}^{32} 2$$

Step2: Apply summation formulas

Recall:

• $\sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6}$
• $\sum_{j=1}^{n} j = \frac{n(n+1)}{2}$
• $\sum_{j=1}^{n} c = cn$ (for constant $c$)

For $n=32$:

1. Calculate $4\sum_{j=1}^{32} j^2$:

$$4 \times \frac{32 \times 33 \times 65}{6}$$

1. Calculate $8\sum_{j=1}^{32} j$:

$$8 \times \frac{32 \times 33}{2}$$

1. Calculate $\sum_{j=1}^{32} 2$:

$$2 \times 32$$

Step3: Compute each term

• Term1: $4 \times \frac{32 \times 33 \times 65}{6} = 4 \times 11440 = 45760$
• Term2: $8 \times \frac{32 \times 33}{2} = 8 \times 528 = 4224$
• Term3: $2 \times 32 = 64$

Step4: Combine the terms

$$45760 - 4224 + 64$$

Answer:

$41600$